Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $r = \dfrac{-3x^2 - 54x - 243}{x^3 + 11x^2 + 18x} \div \dfrac{5x - 25}{-x^2 - 2x} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $r = \dfrac{-3x^2 - 54x - 243}{x^3 + 11x^2 + 18x} \times \dfrac{-x^2 - 2x}{5x - 25} $ First factor out any common factors. $r = \dfrac{-3(x^2 + 18x + 81)}{x(x^2 + 11x + 18)} \times \dfrac{-x(x + 2)}{5(x - 5)} $ Then factor the quadratic expressions. $r = \dfrac {-3(x + 9)(x + 9)} {x(x + 9)(x + 2)} \times \dfrac {-x(x + 2)} {5(x - 5)} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac { -3(x + 9)(x + 9) \times -x(x + 2)} { x(x + 9)(x + 2) \times 5(x - 5)} $ $r = \dfrac {3x(x + 9)(x + 9)(x + 2)} {5x(x + 9)(x + 2)(x - 5)} $ Notice that $(x + 9)$ and $(x + 2)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac {3x\cancel{(x + 9)}(x + 9)(x + 2)} {5x\cancel{(x + 9)}(x + 2)(x - 5)} $ We are dividing by $x + 9$ , so $x + 9 \neq 0$ Therefore, $x \neq -9$ $r = \dfrac {3x\cancel{(x + 9)}(x + 9)\cancel{(x + 2)}} {5x\cancel{(x + 9)}\cancel{(x + 2)}(x - 5)} $ We are dividing by $x + 2$ , so $x + 2 \neq 0$ Therefore, $x \neq -2$ $r = \dfrac {3x(x + 9)} {5x(x - 5)} $ $ r = \dfrac{3(x + 9)}{5(x - 5)}; x \neq -9; x \neq -2 $